Question

One percent of all individuals in a certain population are carriers of a particular disease. A diagnostic test for this disease has a 95% detection rate for carriers and a 4% detection rate for non-carriers. Suppose the test for this is applied independently to two different blood samples from the same randomly selected individual.

Hint: Use Notation A= {no disease} A'={disease} B1= {1st test positive} B2={2nd test positive}

a) What is the probability that the first test is positive?

b) What is the probability that at least one of the tests is positive?

c) What is the probability that the selected individual is a carrier if at least one of the tests is positive?

Answer 1

a)

probability that the first test is positive P(B1)=P(have disease)*P(tested positive |have disease)+P(not have disease)*P(tested positive |not have disease)

=P(A)*P(B1|A)+P(A')*P(B1|A')

=0.01*0.95+(1-0.01)*0.04

=0.0491

b)

probability that at least one of the tests is positive P(B1 u B2)=

P(A)*P(B1 u B2|A)+P(A')*P(B1 u B2|A')

=P(A)*(1-P(B1' n B2'|A))+P(A')*(1-P(B1' n B2''A'))

=0.01*(1-0.05*0.05)+(1-0.01)*(1-0.96*0.96)

=0.0876

c)

P(A |B1 u B2) =P(A)*P(B1 u B2|A)/P(B1 u B2) ==0.01*(1-0.05*0.05)/0.0876 =0.1139